Foundation - Tishk International University

Tishk International University /Civil Engineering Department

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Example A square column foundation to be constructed on a sandy soil has to carry an allowable total load of 250 kN. The depth of the foundation will be 0.7 m. Determine the value width of the footing, B, knowing that c=0, � = 30° and � = 18 ��/��. Use �� = 2 Solution �� = �� + �� + 0.5�� For � = 0, �� = �� + 0.5�� From Table 1-1, for � = 20° �� = 22.46 �� = 19.13 For square footing �� = 1.3 �� = 0.8 So, �� = 0.7(18)(22.46) + 0.5(18)(�)(19.13)(0.8) �� = 283 + 137.736 � �� = �� = 141.5 + 68.85 � (�) Also we have �� = �� × � → �� = 150�� (�) ��(�) = ��(�) 250�� = 141.5 + 68.85 � Solve this Equation, B=1.076 m


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Introduction Foundation settlements must be estimated with great care for buildings, bridges, towers, power plants, and similar high-cost structures. For structures such as fills, earth dams, levees, braced sheeting, and retaining walls a greater margin of error in the settlements can usually be tolerated.


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Settlements are usually classified as follows: 1. Immediate settlement • Takes place as the load is applied or within a time period of about 7 days. • For all fine-grained soils including silts and clays with a degree of saturation, S <≈90 percent., and for all coarse-grained soils with a large coefficient of permeability [say, above 10× 10�� m/s. 2. Consolidation settlement • Time-dependent and takes months to years to develop, for most projects occurring in 3 to 10 years. • For all saturated, or nearly saturated, fine-grained soils where the consolidation theory applies. For these soils we want estimates of both settlement and how long a time it will take for most of the settlement to occur. Immediate settlement computations The settlement of the corner of a rectangular base of dimensions B' X L' on the surface of an elastic half-space can be computed from an equation from the Theory of Elasticity [e.g., Timoshenko and Goodier (1951)] as follows:

�� = �� 1− �� . �� . �� where: ��=immediate settlement, ��=width of footing, �= Poisson ratio of soil, ��= modulus of Elasticity of soil, �=number of corners contributing to settlement (see Fig.(below)), at the footing center � = 4 (point f in the figure), at a side � = 2 (for example, points b in the figure), and at footing corner � = 1 (for example, points a in the figure).


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�� is influence factor, and can be obtained from Figure 2.1

Figure 2.1 Influence factor �� for footing at a depth D. Use actual footing width and depth dimension for this D/B ratio. ��is Steinbrenner influence factor, and can be computed as �� = �� + 1− 2�1− � �� where �� and �� are influence factors, which depend on �/�, thickness of stratum �, Poisson's ratio �, and base embedment depth D. selected range of �� and �� is shown in Table (2.1).


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TABLE 2.1 Values of �� and �� (� = �/�� and � = �/�)


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TABLE 2.1(cont.) Values of �� and �� (� = �/�� and � = �/�)


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The settlement equation (�� = �� . �� . ��) is very widely used to compute immediate settlements. And The equation should be used as follows: 1. Make your best estimate of base contact pressure �� . 2. For round bases, convert to an equivalent square. 3. Determine the point where the settlement is to be computed and divide the base so the point is at the corner or common corner of one or up to 4 contributing rectangles. 4. The stratum depth causing settlement is either of the following: a. Depth � = 5� where � = least total lateral dimension of base. b. Depth to where a hard stratum is encountered. Take "hard" as that where �� in the hard layer is about 10��of the adjacent upper layer. 5. Obtain the weighted average �� in the depth � = �. The weighted average can be computed as ��,�� = �� + �� +⋯+ �� as shown below


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Example 2.1 A rigid shallow foundation 1 m× 2 m is shown below. Calculate the elastic settlement at the center of the foundation.

Solution: Given � = 1� �� = 2�, � = 5� (= 5�) The weighted average of ��,�� is ��,�� = �� + �� + �� ,�� = 2 × 10000 + 1 × 8000 + 2 × 120005 = 10400 ��/��


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� = 4 (at the center) � = �� = 21 = 2 � = �� = �(�/2) = 50.5 = 10

From Table 2.1, �� = 0.641 and �� = 0.031 �� = �� + 1− 2�1− � �� = 0.656 To calculate �� , From Figure 2.1, using �� = �� = 1 and �� = �� = 2, � = 0.3 ⇒ �� = 0.68 Now calculate immediate settlement �� = �� 1− �� . �� . ��

�� = 150(0.5) 1− 0.3�10400 (4)(0.656)(0.68) = 0.0117� = 11.7��


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Example 2.2 Estimate the settlement of the raft (or mat) foundation for the "Savings Bank Building" given by Kay and Cavagnaro (1983). Given data are as follows: �� = 134 ��, � × � = 33.5 × 39.5 m, measured ∆= 18.0��. Soil is layered clays with one sand seam from ground surface to sandstone bedrock at - 14 m; mat at - 3 m. �� from 3 to 6 m = 42.5 MPa �� from 6 to 14 m = 60 MPa, �� for sandstone > 500 MPa. Assume � = 0.35 Solution: The weighted average of ��,�� is ��,�� = �� + �� ,�� = 3 × 42.5 + 8 × 6011 = 55 �� From base to sandstone, � = 14− 3 = 11� � = 4 (at the center) � = �� = 39.533.5 = 1.18 � = �� = �(33.5/2) ≈ 0.7

From Table 2.1, �� = 0.0815 and �� = 0.086 �� = �� + 1− 2�1− � �� = 0.121 To calculate �� ,


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From Figure 2.1, using �� = ��.� = 0.09 and ��.��.� = �� = 2, � = 0.3 ⇒ �� = 0.95 Now calculate immediate settlement �� = �� 1− �� . �� . ��

�� = 134(16.75) 1− 0.35�55 × 1000 (4)(0.121)(0.95) = 0.0165� = 16.5�� This estimate is rather good when compared to the measured value of 18 mm.

Foundation - Tishk International University

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